Book A Free Math Class

# Prove That Root 7 Is Irrational — Proof by Contradiction

**TL;DR**  
To prove that root 7 is irrational, assume the opposite — that 7=\( \frac{p}{q} \) in lowest terms — and derive a contradiction from the fact that 7 must then divide both p and q. This article walks the full proof by contradiction, explains the prime-divisibility step it depends on, and shows the common errors.

## What Is an Irrational Number?

An **irrational number** is a real number that cannot be written as \( \frac{p}{q} \) for any two integers p and q with \( q \neq 0 \). Its decimal expansion never terminates and never settles into a repeating block. Values like \( \sqrt{7} \), π, and e are irrational; every integer, every terminating decimal, and every repeating decimal is rational.

The square root of 7 is a natural candidate because 7 is not a **perfect square** — no whole number squares to 7, since \( 2^2 = 4 \) and \( 3^2 = 9 \). To prove \( \sqrt{7} \) is irrational we must rule out _every_ fraction at once, which is exactly what proof by contradiction does.

## Proof That Root 7 Is Irrational

Here is the full argument laid out cleanly in one place, one step per line. The worked examples below unpack each step in more detail.

**Claim:** \( \sqrt{7} \) is irrational.

Assume the opposite — that \( \sqrt{7} \) is rational.

\( 7 = \frac{p}{q}, \quad \gcd(p, q) = 1, \quad q \neq 0 \)

Square both sides.

\( 7 = \frac{p^2}{q^2} \)

Multiply through by \( q^2 \).

\( p^2 = 7q^2 \)

So 7 divides \( p^2 \). Because 7 is prime, 7 divides p.

Write \( p = 7k \).

Substituting:

\( (7k)^2 = 7q^2 \)  
\( 49k^2 = 7q^2 \)

Divide both sides by 7.

\( 7k^2 = q^2 \)

So 7 divides \( q^2 \), and since 7 is prime, 7 divides q as well.

Now 7 divides both p and q, which contradicts \( \gcd(p, q) = 1 \). The assumption is false, so \( \sqrt{7} \) is irrational.

## The General Method — Why It Works for Any Prime

Nothing in the proof used the specific value 7 except the fact: **7 is prime**. That single property drives both divisibility steps through the prime-divisibility rule:

> If a prime r divides n^2, then r divides n.

Swap 7 for any other prime and the argument runs unchanged. The square roots of 2, 3, 5, and every other prime are irrational for exactly this reason.

## Common Mistakes When Proving Root 7 Is Irrational

### Mistake 1: Forgetting the "lowest terms" assumption

Assuming \( 7 = \frac{p}{q} \) without stating that \( \gcd(p, q) = 1 \).

### Mistake 2: Using the rule for a composite number  
Assuming "r divides n^2 implies r divides n" holds for any r.

### Mistake 3: Trying to prove it from the decimal  
Treating a long non-repeating decimal from a calculator as a finished proof.

## Conclusion  
To prove that root 7 is irrational, assume \( 7 = \frac{p}{q} \) in lowest terms and derive a contradiction. The same proof by contradiction works for the square root of any prime, but not for perfect squares like 4.